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XYLENE POWER LTD.

FNR REACTIVITY

By Charles Rhodes, P.Eng., Ph.D.

REACTOR CORE:
In choosing the gap between the steel fuel tubes the issue is one of maximizing the average fuel density in the fuel bundle while not unduely decreasing the liquid sodium circulation and not imposing unreasonable cleanliness restrictions on the liquid sodium pool.

A related issue is that the maximum sodium temperature as liquid sodium passes through the fuel bundle needs to be limited to provide sufficient temperature safety margin at the upper end of the liquid sodium operating temperature range.

Based on all of these issues the center to center spacing between the square lattice fuel tubes was chosen to be:
(5 / 8) inch = 0.625 inch.

This dimensional choice sets the smallest initial intertube gap in the assembly at (1 / 8) inch, so filtering should be used to eliminate particulates larger than (1 / 32) inch in longest dimension.

Each fuel tube position has associated with it a reactor top surface area of about 1 tube per [0.625^2 inch^2]
= 1 tube / .3906 inch^2

The cross sectional area initially occupied by each fuel tube is:
Pi (.25 inch)^2 = 0.1963 inch^2

Thus the remaining cross sectional area per core fuel tube initially available for natural convection liquid sodium coolant flow is:
0.3906 inch^2 - 0.1963 inch^2 = 0.1943 inch^2
= 0.1943 inch^2 X (.0254 m / inch)^2
= 1.2535 X 10^-4 m^2

Note that initially the cross sectional area of a fuel tube OD and the cross sectional area of a flow channel are almost equal.

Note that a scale diagram shows that initially the flow channel wall area is almost equal to the OD wall area of a fuel tube.

Hence the initial effective cross sectional area for natural circulation through the reactor is:
1.2535 X 10^-4 m^2 / core fuel tube X 476 tubes / bundle X 532 core bundles = 31.7426 m^2

If the fuel tube radius linearly swells by 10% the cross sectional area occupied by each fuel tube becomes:
1.21 X 0.1963 inch^2 = 0.237523 inch^2

Thus the remaining cross sectional area per core fuel tube available for natural convection liquid sodium coolant flow after 10% linear tube swelling is:
0.3906 inch^2 - 0.237523 inch^ = 0.1531 inch^2
= 0.1531 inch^2 X (.0254 m / inch)^2
= 9.8759 X 10^-5 m^2

Fuel tube swelling also increases viscous effects that further reduce primary liqud sodium circulation.

The length of the reactor core perimeter is about:
Pi X 10.4 m = 32.67 m

Hence, prior to fuel tube swelling, the thickness of the radially horizontally moving liquid sodium layer at the active region perimeter is about:
(31.7426 m^2) / (32.67 m = 0.9716 m.
 

PARAMETER DEFINITIONS:
Define:
Ns = average number of sodium atoms / unit volume in reactor core zone
Ni = average number of iron atoms per unit volume in reactor core zone
Nc = average number of chromium atoms per unit volume in ractor core zone
Np = average number of plutonium atoms per unit volume in the reactor core zone with control bundle inserted at start
Npw = averge number of plutonium atoms per unit volume in reactor core zone with contro bundle withdrawn
Nz = average number of zirconium atoms per unit volume in the reactor core zone
Nu = average number of U-238 atoms / unit volume in reactor core zone with control bundle inserted
Nuw = average number of U-238 atoms / unit volume in reactor core zone with control bundle wiothdrawn
Nf = average number of fission product atoms / unit volume in reactor core at end of fuel bundle life with control bundle inserted
Rhos = mass density of liquid sodium = .927 gm / cm^3
Rhoi = mass density of iron = 7.874 gm / cm^3
Rhou = mass density of U-238 = 19.1 gm / cm^3
Rhop = mass density of Pu-239 = 19.8 gm / cm^3
Rhoc = average mass density of control rod slug material Aws = atomic weight of sodium = 23
Awi = atomic weight of iron = 55.845
Awc = atomic weight of Chromium = __________
Awu = atomic weight of uranium = 238
Awz = atomic weight of zirconium = 91.22
Awc = atomic weight of control rod atoms
Av = Avogadro's Number = 6.023 X 10^23 atoms / mole

Define:
Np = number of fissionable plutonium atoms per unit volume
Gu = number of neutrons emitted per average U-235 atomic fission = 2.6
Gp = number of neutrons emitted per average Pu-239 atomic fission = 3.1
Ge = number of neutrons emitted per average atomic fission that are absorbed by breeding
Data from Kaye & Laby for a FNR core:
Sigmaas = fast neutron absorption cross section of sodium = 0.0014 b
Sigmass = fast neutron scatter cross section of sodium = 2.62 b
Sigmaai = fast neutron absorption cross section of iron = .0086 b
Sigmasi = fast neutron scatter cross section of iron = 4.6 b
Sigmaac = fast neutron absorption cross section of chromium = ___
Sigmasc = fast neutron scatter cross section of chromium = _____
Sigmaaf = fast neutron absorption cross section of fission products = ____ b
Sigmaau = fast neutron absorption cross section of U-238 = 0.25 b
Sigmasu = fast neutron scatter cross section of U-238 = 9.4 b
Sigmaap = fast neutron absorption cross section of plutonium-239 = 0.040 b
Sigmaaz = fast neutron absorption cross section of zirconium = 0.0066 b Sigmafp = fast neutron fission cross section of plutonium-239 = 1.70 b
Sigmafu = fast neutron fission cross section of uranium-238 = 0.041 b
 

CORE CRITICALITY:
In order for an FNR to operate there must be a self sustaining chain reaction. This requirement imposes a constraint on the average atomic density of Pu-239 in the reactor core as compared to the average atomic density of U-238, Na-23, Fe, Cr and probability of neutron leakage out the surface of the active core zone.

A FNR inherently operates at the threshold of criticality.

Assume that to support reactor criticality fraction Fp of plutonium fission neutrons remain in the core zone and fraction (1 - Fp) of the plutonium fission neutrons diffuse into the blanket zone. Consider a stream of neutrons following a path of length X within the core zone. The initial flow of neutrons is Fo. Then:
F(X) = Fo Exp{- [(Sigmaas Ns) + (Sigmaai Ni) + (Sigmaac Nc) + (Sigmaau Nu) + (Sigmaaz Nz) + (Sigmaaf Nf) - (Ln(Gp Fp) Sigmafp Np) - (Gu Sigmafu Nu)]}

At the threshold of criticality:
- [(Sigmaas Ns) + (Sigmaai Ni) + (Sigmaac Nc) + (Sigmaau Nu) + (Sigmaaz Nz) + (Sigmaaf Nf) + (Sigmaap Np) - (Ln(Gp Fp) Sigmafp Np) - (Ln(Gu) Sigmafu Nu)]} = 0
or
(Ln(Gp Fp) Sigmafp Np - (Sigmaap Np)) + (Ln(Gu) Sigmafu Nu)] = [(Sigmaas Ns) + (Sigmaai Ni) + (Sigmaac Nc) + (Sigmaau Nu) + (Sigmaaz Nz) + (Sigmaaf Nf)]
or
Np = [(Sigmaas Ns) + (Sigmaai Ni) + (Sigmaac Nc) + (Sigmaau Nu) + (Sigmaaz Nz) + (Sigmaaf Nf) - (Ln(Gu) Sigmafu Nu)] / [(Ln(Gp Fp) Sigmafp) - (Sigmaap)]
or
(Np / Nu) = [(Sigmaas Ns / Nu) + (Sigmaai Ni / Nu) + (Sigmaac Nc / Nu) + (Sigmaau) + (Sigmaaz Nz / Nu) + (Sigmaaf Nf / Nu) - (Ln(Gu) Sigmafu)] / [(Ln(Gp Fp) Sigmafp) - (Sigmaap)]

This equation sets a lower limit on the Pu concentration in the core fuel rods.
 

EVALUATION OF PARAMETERS:
From the web page titled FNR Fuel Rods:
Mass Mu of U-238 in each core fuel rod is:
Mu = .7 (0.287252 kg)
= .20107 kg

Mass of Pu in each core fuel rod is:
Mp = 0.2 (0.287252 kg)
= 0.05745 kg

Mass Mz of Zr in each core fuel rod is:
Mz = 0.1 (0.287252 kg)
= 0.0287252 kg

From the web page titled: FNR FUEL BUNDLE:
Steel cross sectional area in core zone = 24.51%
Core rod cross sectional area in core zone = 15.25%
Sodium cross sectional area in core zone = 60.23%
Area of core zone = 0.160 m^2
Length of Core zone = 0.35 m

Each core zone of a fuel bundle has volume Vcz of: Vcz = 0.400 m X 0.400 m X 0.35 m
= 0.0560 m^3

Within core zone volume Vcz are uranium-plutonium-zirconium fuel rods with a uranium-plutonium-zirconium volume of:
0.1525 Vcz

Within volume Vcz are steel components with a volume of:
0.2451 Vcz

Within volume Vcz is sodium with a volume Vs of:
Vs = 0.6023 Vcz

At the end of a fuel bundle life as set by fuel tube material swelling:
Nf = 2 (0.59 Np)
= 2 (0.59) (Np / Nu) Nu
= 2 (0.59) (2 / 7) Nu
= 0.33714 Nu

Recall that:
Mass Mu of U-238 in each core fuel rod is:
Mu = .7 (0.28725 kg / core rod)
= 0.201075 kg

Mass of Pu in each core fuel rod is:
Mp = 0.2 (0.28725 kg / core rod)
= 0.05745 kg

Mass Mz of Zr in each core fuel rod is:
Mz = 0.1 (0.28725 kg / core rod kg)
= 0.028725 kg / core rod

Mass of iron + chromium in each fuel bundle is:
0.2451 (Vcz)

Ns = (0.6023 Vcz) X (927 kg / m^3) X (1 mole / .023 kg) X Av] / Vcz
= [(0.6023) X (927 kg / m^3) X (1 mole / .023 kg) X Av]

Nu = [476 core rods X 0.201075 kg / core rod X (1 mole / .238 kg) X Av] / (Vcz)

(Ns / Nu) = [(0.6023) X (927 kg / m^3) X (1 mole / .023 kg) X Av] / [476 core rods X 0.201075 kg / core rod X (1 mole / .238 kg) X Av] / (Vcz) = [(0.6023) X (927 kg / m^3) X (1 / .023 kg)] / {[476 core rods X (0.201075 kg / core rod) X (1 / .238 kg)] / (0.0560 m^3)}
= 3.38037

Nz / Nu = (Mz / Mu) (Awu / Awz)
= (0.1 / 0.7) (238 / 91.22)
= 0.3727

Similarly:

(Ni + Nc)= [0.2451 Vcz X Rhoi X (1 mole / Awi) X Av] / (Vcz)
= [0.2451 X Rhoi X (1 mole / 0.055845 kg) X Av]

Hence:
[(Ni + Nc) / Nu] = {[0.2451 X Rhoi X (1 mole / 0.055845 kg) X Av]}
/ [476 core rods X 0.201075 kg / core rod X (1 mole / .238 kg) X Av] / (Vcz)
 
= {[ 0.2451 X 7874 kg / m^3 X (1 mole /0.055845 kg)] [0.0560 m^3]}
/ [476 core rods X 0.201075 kg / core rod X (1 mole / .238 kg)])
 
= 4.8123
 

NUMERICAL EVALUATION:
Recall that:
(Np / Nuc) = [(Sigmaas Ns / Nuc) + (Sigmaai Ni / Nuc) + (Sigmaac Nc / Nuc) + (Sigmaau) + (Sigmaaz Nz / Nuc) + (Sigmaaf Nf / Nuc) - (Ln(Gu) Sigmafu)] / [(Ln(Gp Fp) Sigmafp) - (Sigmaap)]
= [0.0014 b (3.38037) + 0.0086 b (4.8123) + (0.25 b) + 0.0066 b (0.3727) + (Sigmaaf Nf / Nu) - (0.9555) 0.041 b] / [(Ln(3.1 Fp) 1.70 b) - (0.040 b)]
= [.0047325 + .041386 + 0.25 + 0.0024598 + (Sigmaaf Nf / Nuc-b) - 0.03917] / [1.70 b Ln(3.1 Fp) - 0.040]
= [0.2594 + (Sigmaaf Nf / Nuc-b)] / [1.70 Ln(3.1 Fp) - 0.040]
 

Define:
At start of fuel cycle:
Np = Npi
Nuc = Nuci
Npi / Nuci = 0.2 / 0.7 = 0.2857

Assume 15% burnup. Then at the end of the fuel cycle the cumulative number of fissions is 0.75 Npi.

In an element of time dT when (Sigmafp Np) Pu atoms fission (Sigmaau Nuc) U-238 atoms capture neutrons and form new Pu atoms. Thus:
dNuc = - Sigmaau Nuc dT
and
dNp = - Sigmafp Np dT + Sigmaau Nuc dT
or
dNp / dT = - Sigmafp Np + Sigmaau Nuc
or
dNp / (dNuc / - Sigmaau Nuc) = - Sigmafp Np + Sigmaau Nuc
or
dNp / dNuc = (- Sigmafp Np + Sigmaau Nuc) / (- Sigmaau Nuc)
or
dNuc = dNp (- Sigmaau Nuc) / (- Sigmafp Np + Sigmaau Nuc)

The number of fissions in time dT is:
Sigmafp Np dT = Sigmafp Np dNuc / (- Sigmaau Nuc)

Hence:
0.75 Npi = Integral from Np = Npi to Np = Npf of:
{Sigmafp Np dNuc / (- Sigmaau Nuc)}
or
0.75 Npi = Integral from Np = Npi to Np = Npf of:
{Sigmafp Np / (- Sigmaau Nuc)}{dNp (- Sigmaau Nuc) / (- Sigmafp Np + Sigmaau Nuc)
or
0.75 Npi = Integral from Np = Npi to Np = Npf of:
{Sigmafp Np}{dNp / (- Sigmafp Np + Sigmaau Nuc)
or
0.75 Npi = Integral from Np = Npi to Np = Npf of:
{dNp / (- 1 + [(Sigmaau Nuc) /(Sigmafp Np)] )
OR
0.75 Npi = Integral from Np = Npi to Np = Npf of:
{- dNp / ( 1 - [(Sigmaau Nuc) /(Sigmafp Np)] )

To a first approximation:
(- dNp) = (Npi - Npf)

Solution is about: Npi - Npf ~ 0.375 Npi
 

PRACTICAL REACTOR:
Try (Np / Nuc) = (2 / 7) = 0.2857 as in EBR-2. Then:
Ln(3.1 Fp) = {{[0.2594 + (Sigmaaf Nf / Nuc-b)] / (Np / Nuc)} + 0.040} / 1.70
= Ln(3.1 Fp) = {{[0.2594 + (Sigmaaf Nf / Nuc-b)] / (0.2857)} + 0.040} / 1.70
= {[0.9479 + 3.50 (Sigmaaf Nf / Nuc-b)} / 1.70

At the start of a fuel cycle Nf = 0 giving:
Ln(3.1 Fp) = 0.9479 / 1.70 = 0.5576
or
3.1 Fp = Exp (0.5576) = 1.746
or
Fp = 1.746 / 3.1 = 0.5633

At the end of the fuel cycle with 15% burnup:
(Nf / Nuc) ~ (0.3 / 0.65) = 0.4615
and
Np / Nuc = (0.125 / 0.65) = 0.1923
giving:
Ln(3.1 Fp) = {{[0.2594 + (Sigmaaf Nf / Nuc-b)] / (Np / Nuc)} + 0.040} / 1.70
= {{[0.2594 + 0.4615 (Sigmaaf / b)] / 0.1923} + 0.040} / 1.70
= {1.3889 + 2.4 (Sigmaaf / b)} / 1.70
= 0.817 + 1.4117 (Sigmaaf / b)

Hence:
3.1 Fp = Exp[ 0.817 + 1.4117 (Sigmaaf / b)]
or
Fp = Exp[ 0.817 + 1.4117 (Sigmaaf / b)] / 3.1
which for small values of Sigmaaf gives:
Fp = 2.263 / 3.1 = 0.7302

Thus to realize a long fuel cycle time the reactor must be capable of achieving Fp values as high as ~ 0.75

Hence the core rod length will likely need to be extended. The full core rod length is only used near the end of the fuel cycle. Early in the fuel cycle the active core length is only about 60% of the core rod length.

If the core rods are made too long there will be a problem of too much spontaneous reactivity in the active fuel bundles.

The major issue is ensuring that the various shutdown systems work as intended.

In order to control the nuclear reaction full withdrawal of the control portion must render a fuel bundle subcritical under all conditions. Fully inserting the control portion should make the fuel bundle critical provided that 8 adjacent active fuel bundle contol portions are also inserted.
 

NEUTRON DIFFUSION:
In order to sustain the nuclear chain reaction in the reactor core while controlling the reaction by thermal expansion the mean fast neutron travel along the Z axis between successive Pu-239 fissions must be greater than the core zone height. On the web page_________it was determined that a practical core zone height is about 0.35 m.

The neutron path between successive fissions is inversely proportional to the average fissionable fuel density in the reactor. Hence a major issue in FNR design is maximizing the ratio of metallic core fuel volume to total reactor core volume while maintaining adequate cooling and while minimizing unwanted neutron absorption.

REACTIVITY MODULATION VIA TEMPERATURE DEPENDENT DIFFUSION:
CONSIDER THE DISTANCE Lsc TRAVELLED BY A NEUTRON IN A CORE ZONE BETWEEN SUCCESSIVE SCATTERS:
Lsc = 1 / [Ns Sigmass + Ni Sigmasi + Nu Sigmasu]
= 1 / {Nu [(Ns / Nu) Sigmass + (Ni / Nu) Sigmasi + (Nu / Nu) Sigmasu]}
= 10^28 (b / m^2) / {(41.435 X 10^26 U atoms / m^3) [(3.86434) (2.62 b) + (3.9815)(4.6 b) + (1) (9.4 b)]
= 100 m / {41.435 [10.12 + 18.3149 + 9.4]}
= 100 m / {41.435 [37.835]}
= 0.063787 m
Note that as Lsc increases when Ns, Ni, Nu decrease. Hence Lsc increases with increasing temperature.
 

CONSIDER THE DISTANCE Lac TRAVELLED BY A NEUTRON IN THE REACTOR CORE ZONE BEFORE ABSORPTION:
Lac = 1 / [Ns Sigmas + Ni Sigmaai + Nu Sigmaau + Nu Sigmafu + Np Sigmaap + Np Sigmafp]
= 1 / {Nu [(Ns / Nu) Sigmaas + (Ni / Nu) Sigmaai + (Nu / Nu) Sigmaau + (Nu / Nu) Sigmafu + (Np / Nu) (Sigmaap + Sigmafp)]}
= 10^28 (b / m^2) / {(41.435 X 10^26 U atoms / m^3) [(3.86434) (.0014 b) + (3.9815)(.0086 b) + .25 b + .041 b + (0.2857)(.040 + 1.70)]
= 100 m / {41.435 [.00541 + .03424 + .25 + .041 + .497118]}
= 100 m / {41.435 [.82776]}
= 2.915 m
Note that Lac increases when Ns, Ni, Nu, Np decrease. Hence Lac increases with increasing temperature.
 

Hence the number of neutron scatters before absorption is:
(Lac / Lsc) = 2.915 m / 0.063787 m
= 45.71

Since neutron scattering takes place in a 3 dimensional random walk the distance travelled by a neutron along a single axis before absorption is:
[(La / Ls)^0.5 X (Ls / 3^0.5)]
= [(La Ls) / 3]^0.5 = [(2.915 m X 0.063787 m) / 3]^0.5
= .2489 m

Note that diffusion distance [(La Ls) / 3]^0.5 increases with increasing temperature.

Note that by comparison the core zone height is 0.35 m and the adjacent blanket thickness is 1.20 m. The diffusion distance is more than half the core zone height. Hence about half of the fission neutrons diffuse out of the core zone into the blanket zone and do not contribute to maintenance of the chain reaction. An increase in temperature decreases Ns, Ni, Nu, Np which increases:
[(La Ls) / 3]^0.5 which increases the fraction Fp of neutrons diffusing out of the core zone and hence not supporting the chain reaction. Hence the chain reaction will stop at a sufficiently high temperature.
 

OK TO HERE FIND Lsb AND Lab USE DEDICATED WEB PAGE FOR NEXT SECTION

As indicated at FNR FUEL TUBE WEAR at the end of the fuel tube's useful life when the number of plutonium atom fissions equals 0.59 of the number of plutonium atoms initially present:
(Nf / Nu) = 2 (0.59 Np / Np) (Np / Nu)
= 2 (0.59) (.2857) = 0.3371.

DEFINE Fp:
Fp = fraction of plutonium fission neutrons that react in the core zone. Then the fraction of fission neutrons that diffuse into the blanket zones is: (1 - Fp)

This is the condition that must be satisfied for the chain reaction to run when the control portion is fully inserted in the surround portion. At the beginning of the fuel life the 2nd term is zero.
 

NOW CONSIDER THE CASE OF THE CONTROL PORTION FULLY WITHDRAWN:

This is the condition that must be satisfied for the chain reaction to stop when the control rod is fully inserted. Note that at the beginning of the fuel life the 2nd term is zero.

SUMMARY:
Reactor operation is highly dependent on the correct choice of core region height to give the desired value of Fp.

We must solve the diffusion equationto find the diffusion fluxes of neutrons from the core regioninto the adjacent blanket regions to accurately determine Fp in terms of the core rod length Lc.

If Fp is less than 0.27 the reactor will never run. If Fp is larger than 0.49 the chain reaction cannot be stopped_________.

To find Fp we must solve the equation for the diffusion flux of neutrons into the blanket regions above and below the core region.

Thus to make maximum use of the fuel tube material the fast neutron cross section of the fission products must be less than ________. Based on the Kaye & Laby data it appears for fast neutrons the fission product cross sections are much smaller than __________ barnes so fuel bundle working life is determined by fuel tube swelling rather than by fission product accumulation.
 

SAFETY REQUIREMENT:
Note that for safety there must be absolute certainty that the chain reaction can be shut down, so Fp must be significantly less than 0.49. A safe initial choice is Fp = 0.45
 

THERMAL POWER CONTROL:
In a liquid sodium cooled FNR the primary means of thermal power control is thermal expansion which changes the fraction of fission neutrons that diffuse out of the core zone and hence regulates the chain reaction.
 

FUEL BUNDLE REACTIVITY:
The fuel bundles are engineered such that all core zones are sub-critical with the active fuel bundle control portions fully withdrawn. Ideally if a single active fuel bundle in an array of active fuel bundles with their control portions withdrawn has its control portion inserted that fuel bundle should remain subcritical. This feature is desirable to prevent problems if a single control portion jams in its inserted position.

As the reactor core zone exceeds its design operating temperature its thermal expansion should cause the core zone to drop below criticality.

CORE ZONE HEIGHT:
On average each U-235 fission produces 2.6 neutrons and each Pu-239 fission produces 3.1 neutrons. In order for criticality to be maintined one of these fission neutrons must be captured by a plutonium atom. That capture must happen before the neutron leaves the core zone.

Once a neutron leaves the core zone the probability of it being captured by U-238 within the adhacent blanket zone must be vey high.

In a true breeder started with U-235 there is very little margin for neutron loss. Plutonium is a much more practical start fuel.

CRITICALITY MAINTENANCE:
Maintenance of criticality in the core zone requires a minimum core zone size together with a minimum plutonium density within the core zone. From each fission the chain reaction must use at least one neutron to sustain the chain reaction and will likely lose an additional 0.5 neutrons to U-238 absorption in the core. Hence roughly speeking, if La is the average total neutron travel distance from emission to fission:
_______

However, in travelling distance La the neutron goes through about 50 scattering events so its linear travel distance is not that large.

BREEDING CONDITION:
The probability of surplus neutron capture by the surrounding U-238 needs to be at least 95%.

Let Lb = thickness of the blanket surrounding the reactor core.

The blanket is composed of fuel rods that do not contain fissionable material. Then the required minimum blanket thickness Lb is given by:
Lb = 3 /(Sigmaab Nub)
where Nub = average uranium atomic density in the blanket.

FIX FROM HERE ONWARD

Nub =
[(3 X 476) blanket rods X (0.3112 kg / blanket rod) X (238 kg uranium / 260 kg blanket rod) X (6.023 X 10^23 atoms uranium / 0.238 kg uranium) / 0.2400 m^3
= 4.9201 X 10^27 atoms / m^3

Data from the UK Nuclear Data Library indicates that in the blanket the transport cross section is:
Sigmaab = 10.3 b. Hence the minimum required blanket thickness Lb is given by:
Lb = 3 /(Sigmaab Nub)
= 3 / {10.3 X 10^-28 m^2 X 4.9201 X 10^27 atoms / m^3}
= 0.592 m

I have provided for a blanket that is 1.2 m thick to ensure completeness of neutron absorption by U-238.
 

OPERATIONAL NOTE:
The Pu-239 concentration in the blanket may be less than 2% by weight whereas in the core the initial Pu-239 concentration must be about 20% by weight. The Pu-239 concentration in the blanket is gradually bred up to 2%. Then blanket rod reprocessing is used to remove almost 90% of the uranium and zirconium from the blanket fuel rod material to leave behind 20% Pu-239 material which is suitable as fuel for the reactor core.
 

PLUTONIUM DOUBLING TIME:
An issue of great importance in large scale implementation of FNRs is the FNR run time required for one FNR to breed enough excess Pu-239 to allow startup of another identical FNR. This time may be calculated using the approximation that each plutonium atom fission releases of 3.1 neutrons of which 2.5 neutrons are required for sustaining reactor operation leaving 0.6 neutrons for breeding extra Pu-239. Thus one atom of Pu-239 has to fission to form 0.6 atoms of extra Pu-239.

Hence the plutonium doubling time, which the time required to double the available amount of plutonium via breeding within the FNR is:
(10 / 6) X (time to consume the initial Pu supply)
= [(10 / 6) X (1 cycle time )] / 0.59
= 2.825 (1 fuel cycle time)
= 2.825 X (47.78 years)
= 136 years

This is the time required for one FNR to form enough excess Pu to allow starting another FNR. Clearly this doubling time is too long to enable rapid deployment of FNRs.

With large scale implementation of FNRs the available supply of plutonium and trans uranium actinides will soon be exhausted. Hence the issue of the Pu-239 doubling time physically constrains the rate of growth of the FNR fleet.

Thus FNRs are viable for disposing of transuranium actinides but due to the Pu-239 doubling time will not in the near future provide enough power capacity for complete displacement of fossil fuels.
 

This web page last partially updated December 30, 2017

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